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where hpi indicates the average of p. Note that in addition to P1 and P2 , we need to take into account a third term hi1 i2 i which, depending on !1 and !2 , may or may not be zero. (a) By applying superposition in the phasor domain we nd the current in the coil (see Example 9.7). I1 P1 V1 5 0:35 Z1 10 j10 458 A; i1 0:35 cos 2000t 458

4 3 2 x (9 81 30 4 3 3 x (9 81 30 4 3 4 x (9 81 30 4 3 5 x (9 81 30

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Reliability is the probability of not failing in a speci ed length of time. This is usually denoted by R(n), where n is the number of time units. If the time unit is days, then R(1) is the probability of not failing in 1 day. The probability of failing in a speci ed length of time is 1 minus the reliability for that length of time (F n 1 R n ). Software reliability is a measure of how often the software encounters a data input or other condition that it does not process correctly to produce the right answer. Software reliability is not concerned with software wearing out. A better analogy for software failures is picking marbles out of a bag or throwing darts blindfolded at balloons on a wall.

2 RI1 10 0:352 0:625 W 2 2 V 10 608 I2 2 0:45 3:48 A; i2 0:45 cos 4000t 3:48 Z2 10 j20 2 RI2 10 0:452 1W 2 2 i i1 i2 0:35 cos 2000t 458 0:45 cos 4000t 3:48

E(X2) E(X3) E(X4)

8 3(3) 5

P2

11 25

CHAP. 10]

s2 8 3 2 5 32 875

Characteristic functions 3.20. Find the characteristic function of the random variable X of Problem 3.17.

If an error happens every 2 days, then the instantaneous error rate would be 0.5 errors per day. The error rate is the inverse of the time between errors (inter-error time). The error rate can be used as an estimate of the probability of failure, F(1). Unless we know some trend, the best estimate of the short-term future behavior is the current behavior. So if we nd 20 errors on one day, our best estimate for the next day is 20 errors.

In this case hi1 i2 i 0 because hcos 2000t 458 cos 4000t 3:48 i 0. applies and P P1 P2 0:625 1 1:625 W. (b) The current in the coil is i 0:61 cos 2000t 1358 (see Example 9.7). coil is P RI 2 =2 5 0:61 2 1:875 W. Note that P > P1 P2 . (c) By applying superposition in the time domain we nd i1 0:35 cos 2000t 458 ; P1 0:625 W p i2 0:41 cos 1000 2t 35:38 ; P2 0:833 W

The characteristic function is given by E(eivX ) 1 eiv(1) 2

p i i1 i2 ; P hRi2 =2i P1 P2 1:44hcos 2000t 458 cos 1000 2t 35:38 i p The term hcos 2000t 458 cos 1000 2t 35:38 i is not determined because a common period can t be found. The average power depends on the duration of averaging.

1) 1

216 8 35 5

resistor. Find and plot (a) v across the series RC combination and (b) the instantaneous power p entering RC. (c) Compare the results with Examples 10.1 and 10.2.

8 2 6(3) 5

(a) Referring to Fig. 10-2(a), in one cycle of the current the voltages are 1V 0 < t < 1 ms vR 1 V 1 < t < 2 ms t 2000t V 0 < t < 1 ms 1 i dt vC C 0 4 2000t V 1 < t < 2 ms 1 2000t V 0 < t < 1 ms v vR vC [See Fig. 10-13 a 3 2000t V 1 < t < 2 ms (b) During one cycle, pR Ri2 1 mW 2000t mW 0 < t < 1 ms pC vC i 2000t 4 mW 1 < t < 2 ms 1 2000t mW 0 < t < 1 ms p vi pR pC 2000t 3 mW 1 < t < 2 ms (c)

8 4 3 5

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